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Math Help - Can

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Can

    someone give me a quick overview of how to solve congruences? say 3x=24 mod blah blah you get it...I would really appreciate it...I already know how but I am double checking my knowledge base!
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  2. #2
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    Say 3x\equiv 24 (\bmod 60). Note that \gcd(3,60)=3 and 3|24, so the congruence is solvable and has exactly 3 different solutions (up to congruence). If you can find one solution x_0, the others are x_0 + \frac{60}{3} and x_0 + \frac{60}{3}\cdot 2. So it just remains to find x_0. So for example, x_0=8 works. Now you can find everything else.
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